Number of molecules in a volume of 4, cm^{3} | vedclass

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(C) The pressure $P$ is given by $P = h ho g$,where $h = 2\, cm$,$ho = 13.6\, g/cm^{3}$,and $g = 980\, cm/s^{2}$.$P = 2 	imes 13.6 	imes 980 = 2

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$4.0 	imes 10^{18}$

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(C) The pressure $P$ is given by $P = h ho g$,where $h = 2\, cm$,$ho = 13.6\, g/cm^{3}$,and $g = 980\, cm/s^{2}$.$P = 2 	imes 13.6 	imes 980 = 26656\, dyne/cm^{2}$.The volume $V = 4\, cm^{3}$.The mean kinetic energy of a monoatomic gas molecule is given by $E = \frac{3}{2} kT = 4 	imes 10^{-14}\, erg$.From the ideal gas law,$PV = NkT$,where $N$ is the number of molecules.We can write $N = \frac{PV}{kT}$.Since $kT = \frac{2}{3} E = \frac{2}{3} 	imes 4 	imes 10^{-14} = \frac{8}{3} 	imes 10^{-14}\, erg$.Substituting the values: $N = \frac{26656 	imes 4}{\frac{8}{3} 	imes 10^{-14}} = \frac{106624 	imes 3}{8 	imes 10^{-14}} = 13328 	imes 3 	imes 10^{14} = 39984 	imes 10^{14} \approx 4.0 	imes 10^{18}$.

Number of molecules in a volume of $4\, cm^{3}$ of a perfect monoatomic gas at some temperature $T$ and at a pressure of $2\, cm$ of mercury is close to $?$
(Given,mean kinetic energy of a molecule (at $T$) is $4 \times 10^{-14}\, erg$,$g=980\, cm/s^{2}$,density of mercury $=13.6\, g/cm^{3}$)

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Number of molecules in a volume of $4\, cm^{3}$ of a perfect monoatomic gas at some temperature $T$ and at a pressure of $2\, cm$ of mercury is close to $?$(Given,mean kinetic energy of a molecule (at $T$) is $4 \times 10^{-14}\, erg$,$g=980\, cm/s^{2}$,density of mercury $=13.6\, g/cm^{3}$)